For second-order differental equations with linear __constant coefficients__, we can try the solution \(y = e^{kt}\):

\( ay^{''} + by^{'} + cy = 0 \)

\( ak^2e^{kt} + bke^{kt} + ce^{kt} = 0 \)

\( e^{kt}(ak^2 + bk + c) = 0 \)

The exponential term is never zero, thus the quadratic term wrt \(k\) needs to be 0. This equation is the **characteristic equation**
or **auxilliary equation**:

\( ak^2 + bk + c = 0 \)

Recall that the quanity \(b^2 - 4ac\) is the **determinant**. We can solve for values of \(k\) that make this characteristic equation true.
There are three possible outcomes:

Scenario | Determinant | Solution |

2 Real Roots | \(b^2 - 4ac > 0\) | \( y = c_1 e^{k_1t} + c_2 e^{k_2t} \) |

Repeated Real Roots | \(b^2 - 4ac = 0\) | \( y = c_1 e^{k_1t} + c_2 te^{k_1t} \) |

Complex Roots | \(b^2 - 4ac < 0\) | \( y = e^{\alpha t}(c_1 \cos {\beta t} + c_2 \sin {\beta t})\) |

Solve the DE \( y^{''} - 16y^{'} + 48y = 0 \)

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Solve the DE \( y^{''} - 10 y^{'} + 25y = 0 \)

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Solve the DE \( y^{''} + 25y = 0 \)

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